Solve the following : Find the area of the region bounded by the parabola y2 = x and the line y = x in the first quadrant. - Mathematics and Statistics

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Sum

Solve the following :

Find the area of the region bounded by the parabola y2 = x and the line y = x in the first quadrant.

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Solution

To obtain the points of intersection of the line and the parabola, we equate the values of x from both the equations.

∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1). Required area:area of the region OCABO 
= area of the region OCADO – area of the region OBADO
Now, area of the region OCADO
= area under the parabola y2 = x i.e. y = `± sqrt(x)` (in the first quadrant) between x = 0 and x = 1

= `int_0^1 sqrt(x)*dx`

= `[(x^(3/2))/(3/2)]_0^1`

= `(2)/(3) xx (1 - 0)`

= `(2)/(3)`
Area of the region OBADO
= area under the line y = x between x  0 and x = 1

= `int_0^1x*dx`

=`[x^2/2]_0^1`

=`(1)/(2) - 0`

= `(2)/(3)`
∴ required area = `(2)/(3) - (1)/(2)`

= `(1)/(6)"sq unit"`.

Concept: Area Bounded by the Curve, Axis and Line
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Chapter 5: Application of Definite Integration - Miscellaneous Exercise 5 [Page 190]

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Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 5 Application of Definite Integration
Miscellaneous Exercise 5 | Q 2.06 | Page 190
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